101. A chemistry record gives a liquid volume as \(35\,mL\), a sample mass as \(2.5\,g\), and an amount as \(0.10\,mol\). The best interpretation is:
ⓐ. All three values measure the same physical quantity
ⓑ. They mean volume, mass, and amount
ⓒ. \(mL\) measures mass, while \(g\) measures volume
ⓓ. \(mol\) is a unit of temperature used only for gases
Correct Answer: They mean volume, mass, and amount
Explanation: The unit \(mL\) is commonly used for liquid volume in chemistry. The unit \(g\) is commonly used for mass, especially for laboratory-scale samples. The unit \(mol\) measures amount of substance and is essential for counting chemical entities indirectly. These units cannot be interchanged just because they appear in the same record. Unit meaning decides what kind of quantity is being measured before any calculation begins.
102. A measuring cylinder is marked in \(cm^3\), but the solution bottle label uses \(mL\). For ordinary chemistry volume measurements, the useful relation is:
ⓐ. \(1\,mL=1\,cm^3\)
ⓑ. \(1\,mL=10\,cm^3\)
ⓒ. \(1\,cm^3=1000\,mL\)
ⓓ. \(1\,mL=1\,kg\)
Correct Answer: \(1\,mL=1\,cm^3\)
Explanation: The units \(mL\) and \(cm^3\) are commonly treated as equal volume units in chemistry. Thus \(1\,mL\) corresponds to \(1\,cm^3\). This relation is useful when measuring liquids with laboratory glassware. It does not connect volume directly with mass, so \(1\,mL\) should not automatically be called \(1\,g\) unless density information supports that for a specific substance. The equality is between two volume units, not between volume and mass.
103. A solution has volume \(0.250\,L\). What is this volume in \(mL\)?
ⓐ. \(25\,mL\)
ⓑ. \(250\,mL\)
ⓒ. \(2500\,mL\)
ⓓ. \(0.250\,mL\)
Correct Answer: \(250\,mL\)
Explanation: \( \textbf{Known relation:} \) \(1\,L=1000\,mL\).
\( \textbf{Given volume:} \) \(0.250\,L\).
\( \textbf{Required unit:} \) \(mL\).
\( \textbf{Conversion setup:} \) multiply by the unit factor \(\frac{1000\,mL}{1\,L}\).
\[
0.250\,L\times\frac{1000\,mL}{1\,L}=250\,mL
\]
\( \textbf{Unit cancellation:} \) \(L\) cancels and \(mL\) remains.
\( \textbf{Final answer:} \) \(0.250\,L=250\,mL\).
A value such as \(25\,mL\) would come from using \(100\) instead of \(1000\) in the litre-to-millilitre conversion.
104. A student wants to record the volume of a gas collected in a small tube. The tube reading is \(18\,cm^3\). The same volume can be written as:
ⓐ. \(18\,L\)
ⓑ. \(18\,kg\)
ⓒ. \(18\,mol\)
ⓓ. \(18\,mL\)
Correct Answer: \(18\,mL\)
Explanation: The relation \(1\,cm^3=1\,mL\) allows a direct conversion between these two volume units. Therefore, \(18\,cm^3\) is equal to \(18\,mL\). Litre is a larger volume unit, so \(18\,L\) would be much larger than \(18\,cm^3\). Kilogram measures mass and mole measures amount of substance, so neither is a direct volume equivalent. The equality works only because both \(cm^3\) and \(mL\) describe volume.
105. A chemical quantity is written as \(4.0\,g\,mol^{-1}\). The unit most likely belongs to:
ⓐ. Density
ⓑ. Volume
ⓒ. Temperature
ⓓ. Molar mass
Correct Answer: Molar mass
Explanation: The unit \(g\,mol^{-1}\) means grams per mole. Molar mass tells the mass of one mole of a substance, so this is the natural unit for it in many chemistry calculations. Density commonly uses units such as \(g\,cm^{-3}\) or \(kg\,m^{-3}\). Volume uses units such as \(L\), \(mL\), or \(cm^3\), while temperature uses \(K\) or \(^\circ\text{C}\). The expression \(mol^{-1}\) in the denominator is the clue that mass is being related to amount of substance.
106. Use the table below to identify the mismatched quantity-unit pair.
| Row | Quantity | Unit written |
|---|
| P | Mass | \(g\) |
| Q | Volume | \(mL\) |
| R | Amount of substance | \(mol\) |
| S | Temperature | \(g\,mol^{-1}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P is correct because \(g\) is commonly used as a unit of mass in chemistry. Row Q is correct because \(mL\) is a unit of volume. Row R is correct because \(mol\) is the unit of amount of substance. Row S is mismatched because \(g\,mol^{-1}\) is used for molar mass, not temperature. Temperature is commonly expressed in \(K\) in the \(SI\) system or in \(^\circ\text{C}\) in many laboratory contexts.
107. The volume relation \(1\,L=1\,dm^3\) is useful because:
ⓐ. It links two volume units
ⓑ. It proves that litre is a mass unit
ⓒ. It changes every liquid into a gas
ⓓ. It means \(1\,L=1\,mL\)
Correct Answer: It links two volume units
Explanation: Both \(L\) and \(dm^3\) are volume units. The relation \(1\,L=1\,dm^3\) allows volume data to be written in either form without changing the quantity. This is especially useful because solution concentration is often expressed per \(L\), while cubic decimetre notation appears in unit construction. The relation does not involve mass, state change, or amount of substance. It must not be confused with \(1\,L=1000\,mL\), which connects litre with millilitre.
108. A solution volume is \(2.00\,dm^3\). The same volume in \(L\) and \(mL\) is:
ⓐ. \(0.200\,L\) and \(200\,mL\)
ⓑ. \(20.0\,L\) and \(2000\,mL\)
ⓒ. \(2.00\,L\) and \(2000\,mL\)
ⓓ. \(2.00\,L\) and \(2.00\,mL\)
Correct Answer: \(2.00\,L\) and \(2000\,mL\)
Explanation: \( \textbf{Useful relations:} \) \(1\,dm^3=1\,L\) and \(1\,L=1000\,mL\).
\( \textbf{Given volume:} \) \(2.00\,dm^3\).
\( \textbf{First conversion:} \)
\[
2.00\,dm^3=2.00\,L
\]
\( \textbf{Second conversion:} \)
\[
2.00\,L\times\frac{1000\,mL}{1\,L}=2000\,mL
\]
\( \textbf{Unit check:} \) \(dm^3\), \(L\), and \(mL\) are all volume units.
\( \textbf{Final answer:} \) \(2.00\,dm^3=2.00\,L=2000\,mL\).
The option \(2.00\,L\) and \(2.00\,mL\) misses the factor of \(1000\) between \(L\) and \(mL\).
109. Density is obtained by dividing mass by volume. If mass is measured in \(g\) and volume in \(cm^3\), the unit of density becomes:
ⓐ. \(g\,cm^{-3}\)
ⓑ. \(cm^3\,g^{-1}\)
ⓒ. \(g\,mol^{-1}\)
ⓓ. \(mol\,L^{-1}\)
Correct Answer: \(g\,cm^{-3}\)
Explanation: Density is defined as mass per unit volume. When mass is measured in \(g\) and volume is measured in \(cm^3\), the unit becomes \(\frac{g}{cm^3}\), written as \(g\,cm^{-3}\). The unit \(cm^3\,g^{-1}\) would represent volume per mass, which is the reciprocal form. The unit \(g\,mol^{-1}\) belongs to molar mass, and \(mol\,L^{-1}\) belongs to molar concentration. The word "per" in density fixes volume in the denominator.
110. A liquid has mass \(40.0\,g\) and volume \(50.0\,mL\). Taking \(1\,mL=1\,cm^3\), what is its density?
ⓐ. \(0.800\,g\,cm^{-3}\)
ⓑ. \(1.25\,g\,cm^{-3}\)
ⓒ. \(2.00\,g\,cm^{-3}\)
ⓓ. \(90.0\,g\,cm^{-3}\)
Correct Answer: \(0.800\,g\,cm^{-3}\)
Explanation: \( \textbf{Given data:} \) mass \(=40.0\,g\), volume \(=50.0\,mL\).
\( \textbf{Volume conversion:} \) \(1\,mL=1\,cm^3\), so \(50.0\,mL=50.0\,cm^3\).
\( \textbf{Required quantity:} \) density in \(g\,cm^{-3}\).
\( \textbf{Formula:} \)
\[
\rho=\frac{\text{mass}}{\text{volume}}
\]
\( \textbf{Substitution:} \)
\[
\rho=\frac{40.0\,g}{50.0\,cm^3}
\]
\( \textbf{Calculation:} \)
\[
\rho=0.800\,g\,cm^{-3}
\]
\( \textbf{Final answer:} \) The density is \(0.800\,g\,cm^{-3}\).
The value \(1.25\,g\,cm^{-3}\) would come from dividing volume by mass instead of mass by volume.
111. The unit \(mol\,L^{-1}\) is best connected with:
ⓐ. Density
ⓑ. Atomic mass
ⓒ. Time interval
ⓓ. Molarity
Correct Answer: Molarity
Explanation: The unit \(mol\,L^{-1}\) means moles of solute per litre of solution. This is the usual unit for molar concentration or molarity. Density relates mass to volume and may use \(g\,cm^{-3}\) or \(kg\,m^{-3}\). Atomic mass is commonly expressed on the relative atomic mass scale or in \(u\) for atomic-scale mass. The denominator \(L\) in \(mol\,L^{-1}\) shows that a solution volume is involved.
112. The unit for a quantity is constructed as \(\frac{kg}{m^3}\). The quantity is most likely:
ⓐ. Density
ⓑ. Molar mass
ⓒ. Amount of substance
ⓓ. Electric current
Correct Answer: Density
Explanation: The expression \(\frac{kg}{m^3}\) means kilogram per cubic metre. Since density is mass per unit volume, \(kg\,m^{-3}\) is an \(SI\)-based density unit. Molar mass would involve mass per mole, such as \(kg\,mol^{-1}\) or \(g\,mol^{-1}\). Amount of substance is measured in \(mol\), and electric current is measured in \(A\). Unit construction reveals the physical meaning of a derived quantity.
113. A graph is described below.
For one pure liquid at constant temperature, mass is plotted on the y-axis and volume on the x-axis. The graph is a straight line passing through the origin.
The slope of this graph represents:
ⓐ. Density of the liquid
ⓑ. Molar mass of the liquid
ⓒ. Temperature of the liquid
ⓓ. Amount of substance in \(mol\)
Correct Answer: Density of the liquid
Explanation: \( \textbf{Graph axes:} \) y-axis \(=\) mass, x-axis \(=\) volume.
\( \textbf{Slope meaning:} \)
\[
\text{slope}=\frac{\Delta \text{mass}}{\Delta \text{volume}}
\]
\( \textbf{Density relation:} \)
\[
\rho=\frac{\text{mass}}{\text{volume}}
\]
\( \textbf{Reasoning:} \) The slope compares change in mass with change in volume for the same liquid.
\( \textbf{Unit of slope:} \) If mass is in \(g\) and volume is in \(cm^3\), slope has unit \(g\,cm^{-3}\).
\( \textbf{Final answer:} \) The slope represents density.
A straight line through the origin shows that mass is directly proportional to volume for a fixed-density sample.
114. A student writes the unit of molar mass as \(mol\,g^{-1}\). The best correction is:
ⓐ. Molar mass should be written as \(L\,mol^{-1}\)
ⓑ. Molar mass should be written as \(mol\,L^{-1}\)
ⓒ. Molar mass has no unit
ⓓ. Molar mass should be written as \(g\,mol^{-1}\)
Correct Answer: Molar mass should be written as \(g\,mol^{-1}\)
Explanation: Molar mass is mass per mole of a substance. Therefore, the mass unit must be in the numerator and \(mol\) must be in the denominator. The common unit is \(g\,mol^{-1}\). The unit \(mol\,g^{-1}\) would mean moles per gram, which is the reciprocal of molar mass. This distinction matters because reciprocal units change the physical meaning of the quantity.
115. A pressure value is reported as \(1\,atm\), and another as \(1.0\times10^5\,Pa\) approximately. The important unit idea is:
ⓐ. Both are pressure units
ⓑ. \(atm\) is a mass unit and \(Pa\) is an amount unit
ⓒ. Pressure has no unit in chemistry calculations
ⓓ. \(Pa\) is used only for liquid volume
Correct Answer: Both are pressure units
Explanation: Pressure can be expressed using units such as \(Pa\), \(atm\), or \(bar\), depending on the context. The pascal, \(Pa\), is the \(SI\)-based pressure unit, while atmosphere, \(atm\), is common in gas-related chemistry discussions. These units describe the same kind of physical quantity, not mass or amount of substance. Pressure becomes especially important when working with gases. Unit choice must be consistent with the data and formula being used.
116. A formula gives concentration as \(\frac{\text{amount of solute}}{\text{volume of solution}}\). If amount is in \(mol\) and volume is in \(L\), the derived unit is:
ⓐ. \(L\,mol^{-1}\)
ⓑ. \(g\,cm^{-3}\)
ⓒ. \(kg\,m^{-3}\)
ⓓ. \(mol\,L^{-1}\)
Correct Answer: \(mol\,L^{-1}\)
Explanation: The numerator is amount of solute in \(mol\). The denominator is volume of solution in \(L\). Therefore, the unit becomes \(\frac{mol}{L}\), written as \(mol\,L^{-1}\). This unit is used for molar concentration or molarity. The reciprocal \(L\,mol^{-1}\) would mean volume per mole, while \(g\,cm^{-3}\) and \(kg\,m^{-3}\) are density units.
117. The prefix \(kilo\) in a unit such as \(kg\) means:
ⓐ. \(10^3\) times the base unit
ⓑ. \(10^{-3}\) times the base unit
ⓒ. \(10^{-6}\) times the base unit
ⓓ. \(10^{-9}\) times the base unit
Correct Answer: \(10^3\) times the base unit
Explanation: The prefix \(kilo\) represents the factor \(10^3\). Thus \(1\,kg=10^3\,g\) when comparing kilogram and gram as mass units. Prefixes are used to express very large or very small quantities conveniently. The prefix \(milli\) represents \(10^{-3}\), \(micro\) represents \(10^{-6}\), and \(nano\) represents \(10^{-9}\). Confusing \(kilo\) with \(milli\) reverses the size of the unit by a very large factor.
118. A label reports the mass of a sample as \(3.20\,kg\). The same mass in \(g\) is:
ⓐ. \(0.00320\,g\)
ⓑ. \(3.20\,g\)
ⓒ. \(320\,g\)
ⓓ. \(3200\,g\)
Correct Answer: \(3200\,g\)
Explanation: \( \textbf{Known relation:} \) \(1\,kg=10^3\,g=1000\,g\).
\( \textbf{Given mass:} \) \(3.20\,kg\).
\( \textbf{Required unit:} \) \(g\).
\( \textbf{Conversion setup:} \)
\[
3.20\,kg\times\frac{1000\,g}{1\,kg}
\]
\( \textbf{Unit cancellation:} \) \(kg\) cancels and \(g\) remains.
\( \textbf{Calculation:} \)
\[
3.20\times1000=3200
\]
\( \textbf{Final answer:} \) \(3.20\,kg=3200\,g\).
The option \(0.00320\,g\) would come from dividing by \(1000\) instead of multiplying.
119. The prefix \(milli\) is used in \(mg\) and \(mL\). It represents:
ⓐ. \(10^3\)
ⓑ. \(10^{-1}\)
ⓒ. \(10^{-3}\)
ⓓ. \(10^{-9}\)
Correct Answer: \(10^{-3}\)
Explanation: The prefix \(milli\) means one-thousandth of the base unit, or \(10^{-3}\). Thus \(1\,mg=10^{-3}\,g\) and \(1\,mL=10^{-3}\,L\). It does not mean \(10^3\), which belongs to \(kilo\). It also differs from \(deci\), which means \(10^{-1}\), and \(nano\), which means \(10^{-9}\). The same prefix factor applies even when the physical quantity changes from mass to volume.
120. A solution bottle contains \(0.075\,L\) of liquid. What is this volume in \(mL\)?
ⓐ. \(0.075\,mL\)
ⓑ. \(0.75\,mL\)
ⓒ. \(75\,mL\)
ⓓ. \(7500\,mL\)
Correct Answer: \(75\,mL\)
Explanation: \( \textbf{Known relation:} \) \(1\,L=1000\,mL\).
\( \textbf{Given volume:} \) \(0.075\,L\).
\( \textbf{Required unit:} \) \(mL\).
\( \textbf{Use the conversion factor:} \)
\[
0.075\,L\times\frac{1000\,mL}{1\,L}
\]
\( \textbf{Cancel units:} \) \(L\) cancels from numerator and denominator.
\( \textbf{Numerical step:} \)
\[
0.075\times1000=75
\]
\( \textbf{Final answer:} \) \(0.075\,L=75\,mL\).
A result such as \(0.075\,mL\) keeps the number unchanged and ignores the \(10^3\) relation between \(L\) and \(mL\).
